= Mating assortment contrast
{title2=$I=(N_{\mathrm{same}}-N_{\mathrm{cross}})/N$}
For observed pairings between two labels $A,B$, let $N_{\mathrm{same}}=N_{AA}+N_{BB}$ and $N_{\mathrm{cross}}=N_{AB}+N_{BA}$, where the first label denotes the female's group and the second the male's. A <mating assortment contrast> is
$$
I=\frac{N_{\mathrm{same}}-N_{\mathrm{cross}}}{N},\qquad N=N_{\mathrm{same}}+N_{\mathrm{cross}}>0.
$$
It lies between $-1$ and $1$. Positive $I$ describes an excess of same-label pairings over cross-label pairings; whether that is evidence of <assortative mating> depends on the null <mathematical model> and the assay. The four counts form a <contingency table>.
Here is an original scope check. Suppose independently paired females and males have <probabilities> $p_f,p_m$ of carrying label $A$. Then
$$
q_{\mathrm{same}}=p_fp_m+(1-p_f)(1-p_m),\qquad \mathbb E[I]=2q_{\mathrm{same}}-1.
$$
With $p_f=p_m=1/2$, the <expected value> is zero. With $p_f=p_m=4/5$, it is $9/25=0.36$, despite pairing being independent of label. This counterexample concerns unequal pool composition; it does not replace the group introductions and <contingency table> tests in the <dietary divergence and mating isolation in Drosophila pseudoobscura> experiment. Even with balanced pools, finite observed samples fluctuate: a positive $I$ alone is not a <hypothesis test>, much less a universal <species> boundary.
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